Topic 5
Gases
Gases have several characteristics that distinguish them from
liquids and solids. For one, gases can be compressed into smaller
containers or volumes. Gas molecules have plenty of space between
them thereby allowing for compression while solid and liquid
molecules are touching thereby restricting their compressibility.
Gases also can relate pressure (P), volume (V), temperature (T),
and number of mols (n) with a fair degree of accuracy by empirical
relationships.
1
Gas Laws
We will examine the quantitative relationships,
or empirical laws, governing gases.
Basically, we will learn about equations that are
available to calculate P, V, T, & n for gases.
Note: these equations are for gases only!
2
Pressure
What is pressure?
Pressure - force exerted per unit area of surface
by molecules in motion.
P = Force/unit area
1 atmosphere (atm) = 14.7 psi
1 atmosphere = 760 mm Hg = 760 Torr (memorize)
1 atmosphere = 101,325 Pascals
1 Pascal
= 1 kg/m.s2
1 atm = 0.101325 MPa = 1.01325 bar (helpful in HW)
3
Chemist have units of pressure based on the
mercury barometer (mm Hg).
A mercury
barometer
Barometer – device for measuring the
pressure of the atmosphere.
It consists of a glass tube about 1 meter in
length filled with mercury, Hg, and inverted
in a dish of the same liquid metal.
At sea level, the Hg in the tube is at a height
of about 760 mm Hg above the level in the
dish. This height is a direct measure of
atmospheric pressure. 760 mm Hg is
referred to as standard pressure at 25oC.
We also refer to standard pressure as being
equal to 1 atm (1 atm = 760 mm Hg).
A similar process for measuring pressure in
a vessel is done by using a sealed u-type
flask called a manometer.
4
Pressure Conversions
The pressure of gas in a flask is 797.7 mmHg. What is the
pressure in atm?
For this type of problem, you need to perform dimensional
analysis to cancel units.
Note: 760 mm Hg is an exact number and does not affect significant digits in
answer.
HW 37
code: pressure
5
The Empirical Gas Laws
All gases behave quite simply; This allows relationships to be
determined by holding any two physical properties constant (P,
V, T, or n) which leads us to the empirical gas laws. The gas
property of compressibility lead to the discovery of
Boyle’s Law: The volume of a sample of gas at
a given temperature varies inversely with the
applied pressure.
V a 1/P
(at constant moles (n) and T)
meaning, PV = constant
and therefore
P1  V1  P2  V 2
6
A Problem to Consider
V1
P1
A sample of chlorine gas has a volume of 1.8 L at 1.0 atm.
P2
If the pressure increases to 4.0 atm (at constant
temperature), what would be the new volume?
We assume constant number of mols of gas since there is no evidence of gas
escaping. Since we are at constant temperature, it allows us to use Boyle’s
Law; we know the pressure and volume at a given temperature and a new
pressure later on so we can easily solve for the new volume, V2.
P1V1 = P2V2
rearranging to solve for the new volume, V2
Notes: 1.) as long as the pressures or volumes are the same unit, they will cancel
out. 2.) Since the pressure increased and we know the volume will have the
inverse effect, we anticipate the volume to decrease which it does (1.8 L to 0.45 L).
HW 38
code: boyle
7
The Empirical Gas Laws
Temperature also affects gas volume. A gas contracts
when cooled & expands when heated which gets us to
Charles’s Law: The volume occupied by any sample
of gas at constant pressure and mols is directly
proportional to its absolute temperature, meaning Kelvin
scale.
V a Tabs
(at constant n and P)
meaning, V = constant
T
and therefore
V1
T1

V2
T2
where T must
be in K.
8
A Problem to Consider
A sample ofT methane gas that
has a volume of
T
V1
1
2
3.8 L at 5.0°C is heated to 86.0°C at constant pressure.
Calculate its new volume.
We assume constant number of mols of gas since there is no evidence of gas
escaping. Since we are at constant pressure, it allows us to use Charles’s
Law; we know the temperature and volume at a given pressure and a new
temperature later on so we can easily solve for the new volume, V2.
V1
T1

V2
T2
rearranging to solve for the new volume, V2
and using Kelvin temperature we obtain
Notes: 1.) you must use Kelvin temperature: K = oC + 273.15 K
2.) Since the temperature increased and we know the volume will also increase, we
anticipate the volume to increase which it does (3.8 L to 4.9 L).
HW 39
code: charles
9
The Empirical Gas Laws
Amontons’ Law (also know as Gay-Lussac’s Law) : The
pressure exerted by a gas at constant volume and
number of mols is directly proportional to its
absolute temperature, K.
P a Tabs
(at constant n and V)
meaning, P = constant
T
and therefore
P1
T1

P2
T2
where T must
be in K.
10
A Problem to Consider
P1
T1
An aerosol can has a pressure of 1.4 atm at 25°C. What
pressure would it attain at 1200°C, assuming the volume
T2
remained constant?
We assume constant number of mols of gas since there is no evidence of gas
escaping. Since we are at constant volume, it allows us to use Amontons’
Law; we know the temperature and pressure at a given volume and a new
temperature later on so we can easily solve for the new pressure, P2.
P1
T1

P2
T2
rearranging to solve for the new pressure, P2
and using Kelvin temperature we obtain
HW 40
code: temp
Notes: 1.) you must use Kelvin temperature: K = oC + 273.15 K
2.) Since the temperature increased and we know the pressure will also increase,
we anticipate the pressure to increase which it does (1.4 atm to 6.9 atm).
11
The Empirical Gas Laws
Combined Gas Law: In the event that all three
parameters, P, V, and T, are changing, their
combined relationship is defined as follows (at
constant n):
PV a Tabs
( at constant n)
meaning, PV = constant
T
and therefore
P1V1
T1

P2V2
T2
where T must
be in K.
Note: the combined gas law involves all three previous laws discussed and can be
used to solve any problems involving constant n. If any of the variables are
constant, they will simply cancel out in the calculation.
12
A Problem to Consider
V1
T1
A sample of carbon dioxide gas occupies 4.5 L atP30°C and
P1
2
650 mmT2Hg. What volume would it occupy at 800 mm Hg
and 200°C?
We assume constant number of mols of gas since there is no evidence of gas
escaping. Since this problem involves one set of conditions as compared to
another set of conditions, we will use the combined gas law; we know the
temperature, volume, and pressure of the gas and are given a new
temperature and pressure so we can easily solve for the new volume, V2.
P1V1
T1

P2V 2
T2
rearranging to solve for the new volume, V2
and using Kelvin temperature we obtain
HW 41
code: combine
13
The Empirical Gas Laws
Avogadro’s Law: At a given temperature and pressure,
the volume of a gas sample is directly proportional to
the number of moles of gas particles in the gas sample.
Vn
at constant T & P
This implies that if we have twice as many gas particles, the gas would
occupy twice as much volume at the same T & P. Another way of
stating Avogadro’s law is to say: at the same T & P, equal volumes of
any gas contain equal number of particles.
The volume of one mole of any gas is called the molar
gas volume, Vm and has the units of L/mol.
14
The Empirical Gas Laws
Volumes of gases are often compared at standard
temperature and pressure (STP), chosen to be 0 oC (273 K)
and 1 atm pressure.
At STP, the molar volume, Vm, that is, the
volume occupied by one mole of any gas, is
22.4 L/mol
Conversion factor for gases at STP:
1 mol of any gas = 22.4 L
1 mol H2 at STP has volume of 22.4L
1 mol of O2 at STP has a volume of 22.4 L
1 mol of any gas at STP has a volume of 22.4 L
HW 42
code: molar
15
The Ideal Gas Law
From the empirical gas laws, we have shown that
volume varies in proportion to pressure, absolute
temperature, and moles.
V  1/P
V  Tabs
Boyle' s Law
Charles' Law
Vn
Avogadro' s Law
combining gives
V  n Tabs /P
16
The Ideal Gas Law
V  n Tabs /P
This implies that there must exist a proportionality
constant governing these relationships meaning they
must be equal based on some constant
V " R" (
nTabs
P
)
where “R” is the proportionality constant referred
to as the ideal gas constant (independent of gas
species).
17
The Ideal Gas Law
We can calculate R at STP using the ideal gas law
because the only unknown is “R”:
1 mol of any gas has a volume of 22.4 L at 273 K
and 1 atm of pressure
V " R" (
nTabs
P
R

)
rearranging to
R
VP
nT
(22.4 L)(1.00 atm)
(1.00 mol)(273 K)
Latm
0.0821 molK
Note: memorize this common value for R. There are other values of R that
can also be used; however, caution should be taken in the differences in units.
18
The Ideal Gas Law
Thus, the ideal gas equation, is usually
expressed in the following form:
PV  nRT
P is pressure (in atm)
V is volume (in liters)
n is number of atoms (in moles)
R is universal gas constant - 0.0821 L.atm/mol . K
T is temperature (in Kelvin)
Note: R value with correct units to cancel with units given for the other variables.
19
A Problem to Consider
An experiment calls for 3.50 moles of chlorine gas, Cl2.
What volume would this be if the gas volume is measured
at 34°C and 2.45 atm?
Since this problem involves P, V, T, and n, we will use the ideal gas law; we
know the temperature, mols, and pressure of the gas so we can easily solve
for the volume.
PV  nRT
rearranging to solve for the volume
and using Kelvin temperature we obtain
HW 43
code: ideal
20
Molecular Weight Determination
In section 3 we showed the relationship
between moles and mass.
molar mass, Mm

mass,m
moles,n
or
n
m
Mm
21
Molecular Weight Determination
n  Mm
If we substitute
m
into the ideal gas equation PV  nRT , we obtain
PV 
m
( M m )RT
If we solve this equation for the molar mass, we obtain
mRT
Mm 
PV
The procedure for determining the molar mass of a gas
from measurements of m, P, T, and V is known as the
Dumas method.
22
A Problem to Consider
A 9.25 gram298
sample
K of an unknown gas occupied a volume
of 5.75 L at 25°C and a pressure of 1.08 atm. Calculate its
molar mass. Which of the following gases is most likely to
be the unknown gas - N2, O2, or HCl?
mRT
Mm 
PV
Molar mass:
N2 – 28.02 g/mol
Therefore, most likely
unknown gas is HCl.
O2 – 32.00 g/mol
HCl – 36.46 g/mol 23
Density Determination
We can use the Dumas method, also known as the
vapor density method, to calculate molar mass based
on the density of a gas.
If we look again at our derivation of the molar mass equation,
PV  ( Mmm ) RT
we can solve for molar mass and insert density,d = m/V:
mRT dRT
Mm 

VP
P
We can use the above equation to solve for molar mass; however, if we
are at STP, we can simplify things further because we also know that
any gas at STP has a molar volume of 22.4 L/mol meaning
22.4 L V RT
Vm 
 
mol
n
P
therefore,
Mm = d Vm @ STP
24
A Problem to Consider
Calculate the density of ozone gas, O3 (Mm = 48.0 g/mol), at
50°C and 1.75 atm of pressure.
323 K
We must use the Dumas method equation and rearrange to solve for density.
Note: we can’t use the simpler equation because we are not at STP.
dRT
Mm 
P
rearranges to
PM m
d 
RT
Note: typical density units for a gas are g/L
HW 44
25
code: vary
Stoichiometry Problems Involving Gas Volumes
2 KClO 3 (s)  2 KCl(s)  3 O 2 ( g )
0.0100 mol
?L
Suppose you heat 0.0100 mol of potassium chlorate, KClO3, in a test
tube. How many liters of oxygen can you produce at 298 K and 1.02 atm?
This type of problem is similar to other stoichiometry problems except for
the fact that we are looking for the volume of O2 generated instead of mols or
mass. The mols of gas is related to the volume through the ideal gas law.
Stoichiometry:
Ideal gas law (PV = nRT) solving for V:
26
Stoichiometry Problems Involving Gas Volumes
Many air bags are inflated with N2 gas by the following rxn:
6NaN3 (s) + Fe2O3 (s)  3 Na2O (s) + 2Fe (s) + 9N2 (g)
?g
n
How many grams of NaN3 would be needed to provide 75.0 L of
N2 gas at 25oC and 748 mmHg?
298 K
This problem gives us the information needed to calculate the mols of N2 gas to
inflate the air bag by using the ideal gas law. Once we have the mols of N2 gas, we
can calculate the mass of sodium azide needed to produce that amount of gas through
stoichiometry.
Ideal gas law (PV = nRT) solving for n:
Stoichiometry:
HW 45
code: stoich
27
Partial Pressures of Gas Mixtures
Dalton’s Law of Partial Pressures: the sum of
all the pressures of all the different gases in a
mixture equals the total pressure of the mixture.
28
Partial Pressures of Gas Mixtures
The composition of a gas mixture is often
described in terms of its mole fraction.
The mole fraction of a component gas is the
fraction of moles of that component in the total
moles of gas mixture
nA
 A  mole fraction of A 
n tot
note: no units on
mole fraction
Mole fraction can also be calculated through partial
pressure.
29
Partial Pressures of Gas Mixtures
By rearranging
the partial pressure of a component gas, “A”, is
then defined as
PA   A  Ptotal
30
A Problem to Consider
291 K
A 10.0 L flask contains 1.031 g O2 and 0.572 g CO2 gases at 18oC. What
are the partial pressures of O2 and CO2? What is the total pressure?
What is the mole fraction of O2 in the mixture?
We know that each gas can be treated independently by using the ideal gas law
PO2V = nO2RT and PCO2V = nCO2RT and rearranging to solve for pressure.
Total pressure is the sum of all gas pressures:
Mole fraction may be calculated in several ways but the easiest way in this
problem will be to use the partial pressure:
HW 46
31
code: partial
Collecting Gases “Over Water”
A useful application of partial pressures arises
when you collect gases over water.
Basically, a gas displaces water in an amount equal in
volume to the gas (Vgas = VH2O collected).
As gas bubbles through the water, the gas becomes
saturated with water vapor meaning that you must
account for the water vapor pressure in the measured
pressure.
The partial pressure of the water in this “mixture”
depends only on the temperature (vapor pressure of
water which can be looked up in reference books).
32
A Problem to Consider
Suppose a 156 mL sample of H2 gas was
collected over water at 19oC and 769 mm Hg.
What is the mass of H2 collected?
First, we must determine the partial pressure of the hydrogen gas since the
total pressure involves water vapor. Note that the water vapor pressure can be
obtained from water vapor pressure tables in reference books at the correct
temperature which in this case is 19oC (16.5 mm Hg).
PH2 = Ptotal - PH2O = 769 mm Hg – 16.5 mm Hg = 752 mm Hg
33
Suppose a 156 mL sample of H2 gas was collected over water at 19oC and 769
292 K
mm Hg. What is the mass of H2 collected?
Next, we can determine the mols of hydrogen gas through the ideal gas law,
PH2V = nH2RT, by using the partial pressure of hydrogen gas and the volume
of water displaced since it equals the volume of hydrogen gas generated. We
also will need to make some conversions for units can cancel.
Lastly, we need to convert from mols to mass of hydrogen:
HW 47 & 48
code: gas
34
Kinetic-Molecular Theory of gases
A simple model based on the actions of individual atoms used to
explain the behavior of ideal gases. There are five postulates to
this theory:
• Volume of particles can be neglected but volume of container
cannot (smaller container, higher probability of gas molecules
having collisions).
• Particles are in constant random motion; move in straight lines
in all directions and at various speeds (smaller mass, faster it
moves, higher probability of collisions).
• No inherent attractive or repulsive forces
• When molecules collide, the collisions are elastic (total KE
remains constant; there may be a transfer of energy but none
lost).
• The average kinetic energy of a collection of particles is
proportional to the temperature (K) – higher T, greater KE
(higher the temp, faster molecules move, higher probability of
collisions)
Gas pressure is due to collisions of gas particles with a surface (container). If we
add up the forces due to all the collisions of gas particles with the surface 35
(container), and divide the result by the area of the surface, we get the pressure.
Molecular Speeds; Diffusion and Effusion
The root-mean-square (rms) molecular
speed,  – m/s, is a type of average molecular
speed, equal to the speed of a molecule having
the average molecular kinetic energy. It is given
by the following formula:
3RT
v 
Mm
rms
36
Molecular Speeds; Diffusion and Effusion
Diffusion is the transfer of a gas through space or
another gas over time from a region of high
concentration (more crowded) to a region of low
concentration (less crowded)
Effusion is the transfer of a gas through a membrane
or orifice.
The equation for the rms velocity of gases shows the
following relationship between rate of diffusion and
molecular mass (inversely proportional).
Graham’s Law:
1
Rate of diffusion 
Mm
expect larger molecules
(larger Mm) to move
slower and hence, slower
diffusion rates.
37
The rate of diffusion/effusion of molecules from a container
depends on three factors:
1.) cross-sectional area of the hole (the larger it is; the more likely
molecules are to escape)
2.) the number of molecules per unit volume (conc of gas); the
more crowded the molecules, the more likely for the molecules to
diffuse.
3.) the average molecular speed (affected by temp and molar
mass); faster the molecules move (smaller the Mm) and higher the
temperature, the higher probability they will be able to diffuse.
The temp, conc., molar mass, and size of the actual hole involved
in gas molecules escaping affects the rate of diffusion/effusion.
Therefore, if you compare the diffusion/effusion of different gases
that are the same concentration, in the same container, have the
same T & P, then the one factor that will dictate the rate would38
be the molar mass of species.
Molecular Speeds; Diffusion and Effusion
According to Graham’s law, the rate of effusion
or diffusion is inversely proportional to the
square root of its molecular mass. (for gases in
same container at constant T & P). The
following relationship allows for comparison of
gases:
Rate of effusion of gas " A"

Rate of effusion of gas " B"
M m of gas B
M m of gas A
39
A Problem to Consider
How much faster would H2 gas effuse through
an opening than methane, CH4?
We expect hydrogen gas to effuse faster than methane gas because it is
lighter gas so we will put it in the numerator of our comparison equation:
Rate of H 2
M m (CH 4 )

Rate of CH 4
M m (H 2 )
Rate of H 2
16.0 g/mol

 2 .8
Rate of CH 4
2.0 g/mol
So hydrogen effuses 2.8 times faster than CH4
HW 49
code: effuse
40